This web page has every one of the required homework for the product covered in the second exam the the 2nd semester of general Chemistry. The textbook linked with this homework is CHEMISTRY The central Science through Brown, LeMay, et.al. The critical edition I compelled students to buy was the 12th execution (CHEMISTRY The main Science, 12th ed. By Brown, LeMay, Bursten, Murphy and Woodward), but any type of edition of this message will do for this course.

Note: You space expected to go to the end of chapter troubles in your textbook, find similar questions, and work the end those troubles as well. This is just the required perform of problems for quiz purposes. You should additionally study the Exercises within the chapters. The exercises are cleared up examples the the questions at the back of the chapter. The study guide likewise has cleared up examples.

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These are bare-bones questions. The textbook inquiries will have added information that might be useful and also that connects the difficulties to real life applications, many of castle in biology.

Properties of options (Chapter Thirteen)The thickness of ethanol is 0.789 g/mL and also the thickness of water is 1.0 g/mL. A equipment is made whereby 58.3 mL the ethanol, C2H5OH, is dissolved in 500 mL that water. Assume the volumes room additive.What is the mole fraction of ethanol in this solution?Answer
 ( 58.3 mL ) ( 0.789 g 1 mL
)( 1 mole
46 g
) = 1 mole C2H5OH
 ( 500 mL H2O ) ( 1 g 1 mL
)( 1 mole
18 g
) = 27.8 mole H2O

The full moles = 1 mole C2H5OH + 27.8 mole H2O = 28.8 moles

 The mole fraction = mole C2H5OH full moles
= 1 mole C2H5OH
28.8 complete moles
= 0.035

The mole percent would certainly be 3.5%.

 ( 58.3 mL ) ( 0.789 g 1 mL
) = 46 g C2H5OH
 ( 500 mL H2O ) ( 1 g 1 mL
) = 500 g H2O

The full mass = 46 g C2H5OH + 500 g H2O = 546 g

 The mass portion = fixed C2H5OH full mass
= 46 g C2H5OH
546 full grams
= 0.084

The fixed percent would be 8.4%.

 The molarity = mole C2H5OH l of systems
= 1 mole C2H5OH
.5583 l
= 1.79 M What is the molality?Answer The molality =
 moles C2H5OH kg the solvent
= 1 mole C2H5OH
0.5 kg H2O
= 2 m isopropil alcohol in ~ the drug keep is usually 70% isopropyl alcohol, C3H7OH, and 30% water by volume. This synchronizes to 54.94 g C3H7OH and also 30 g H2O in 100 mL. The density of this systems is 0.8494 g/mL. In this difficulty assume water is the solute and the alcohol is the solvent.What is the mole fraction and mole percent of water in this solution?Answer (
 54.94 g C3H7OH ) ( 1 mole C3H7OH 60 g C3H7OH
) = 0.916 mole C3H7OH
 ( 30 g H2O ) ( 1 mole 18 g H2O
) = 1.67 mole H2O

The full moles = 0.916 mole C3H7OH + 1.67 mole H2O = 2.586 moles

 The mole fraction = moles H2O complete moles
= 1.67 mole H2O
2.586 full moles
= 0.646

The mole percent would certainly be 64.6%.

What is the mass portion and fixed percent of water?Answer

The complete mass = 54.94 g C3H7OH + 30 g H2O = 84.94 g

 The mass fraction = fixed H2O full mass
= 30 g C2H2O
84.94 total grams
= 0.353

The massive percent would certainly be 35.3%.

What is the molarity of water?Answer

To get the liters of systems we usage the complete mass and the density. The full mass is 54.94 g to add 30 g or 84.94 g.
 ( 84.94 g solution ) ( 1 mL 0.8494 g solution
) = 100 mL of equipment

 ( 30 g H2O ) ( 1 mole 18 g H2O
) = 1.67 mole H2O

 The molarity = moles H2O together of equipment
= 1.67 mole H2O
0.1 together
= 16.7 M What is the molality of water?Answer The molality =
 mole H2O kg of solvent
= 1.67 mole H2O
0.05494 kg C3H7OH
= 30.4 m What will the melting suggest and the boiling suggest be for a solution where 5 mL that hexane, C6H14, (density = 0.658 g/mL) is inserted into 100 mL of benzene, C6H6 (density = 0.8765 g/mL)? The regular boiling point of benzene is 80.1 °C and also the typical freezing point of benzene is 5.5 °C. For benzene the boiling suggest elevation constant, kb, is 2.53 °C/m and also the freezing allude depression constant, Kf, is 5.12 °C/m.Answer (
 5 mL C6H14 ) ( 0.658 g C6H14 1 mL C6H14
)( 1 mole C6H14
86 g C6H14
) = 0.0383 mole C6H14

 ( 100 mL C6H6 ) ( 0.8765 g C6H6 1 mL C6H6
)( 1 kg C6H6
1000 g C6H6
) = 0.08765 kg C6H6

 The molality = 0.0383 mole C6H14 0.08765 kg C6H6
= 0.437 m

ΔTf = kfm =(5.12 °C/m)(0.437 m) = 2.24 °C

New f.p. = 5.5 °C - 2.24 °C = 3.26 °C

ΔTb = kbm =(2.53 °C/m)(0.437 m) = 1.1 °C

New b.p. = 80.1 °C + 1.1 °C = 81.2 °C How countless liters that ethylene glycol, C2H6O2, need to be included to 3 l of water to make a equipment that will certainly freeze at -20 °C (-4 °F)? The thickness of C2H6O2 is 1.11 g/ml and the density of water is 1 g/ml.AnswerΔT = kfm ⇒ 20 °C = (1.86 °C/m)m ⇒

The molality
 = 20°C 1.86 °C/m
= 10.75 m = 10.75 moles C2H6O2
kg the water

 ( 3 L H2O ) ( 1 kg H2O 1 together H2O
)( 10.75 mole C2H6O2
1 kg H2O
) = 32.25 mole C2H6O2

 ( 32.25 mole C2H6O2 ) ( 62 g C2H6O2 1 mole C2H6O2
)( 1 mL C2H6O2
1.11 g C2H6O2
) = 1800 mL C2H6O2 = 1.8 L list the adhering to aqueous options in the stimulate of enhancing freezing point: 0.5 m Ca(NO3)2, 0.8 m sucrose, 0.6 m LiF.Answer
 ( 0.5 mole Ca(NO3)2 1 kg water
)( 3 moles of corpuscle
1 mole Ca(NO3)2
) = 1.5 m in particles

 ( 0.8 mole sucrose 1 kg water
)( 1 moles of particles
1 mole sucrose
) = 0.8 m in particles

 ( 0.6 mole LiF 1 kg water
)( 2 mole of particles
1 mole LiF
) = 1.2 m in corpuscle

ΔTf = kfm says that the readjust in freezing allude is straight related come the molality that the particles in the solution. The larger the molality, the larger the ΔT and the lower the freeze point. The greatest freezing allude will be the solution with the smallest molality, i m sorry is sucrose. The LiF equipment will it is in next and the Ca(NO3)2 solution will have the shortest freezing point.Nandrolone is one anabolic steroid (a muscle-building chemical) which occurs normally in the person body, however only in small quantities. A 20 g sample the nandrolone was placed in 500 mL the CCl4 (density = 1.59 g/mL) and also the freezing allude of the equipment was found to it is in 1.75 °C lower than the normal freezing point. What is the molecular weight of nandrolone? kf because that CCl4 is 29.8 °C/m. Answer
 The molality = ΔT kf
= 1.75 °C
29.8 °C/m
= 0.0587 m = 0.0587 mole nandrolone
kg of CCl4
(500 mL CCl4)(1.59 g/mL) = 795 g CCl4 = 0.795 kg CCl4

 ( 0.795 kg CCl4 ) ( 0.0587 mole nandrolone 1 kg CCl4
) = 0.04667 mole nandrolone

 The molecular load = 20 g nandrolone 0.04667 mole nandrolone
= 428.5 g/mole (C28H44O3) The partial push of oxygen, O2, in ~ the greatest recorded atmospheric press was 0.225 atm and also the concentration in water under those problems was 6.5 x 10-5 M. What was the concentration in water in ~ the lowest tape-recorded atmospheric pressure when the partial press of oxygen was 0.18 atm? Fish require 1.56 x 10-7 M O2 in order to survive. Carry out they have a high enough concentration in ~ the lowest recorded atmospheric pressure?AnswerUsing Henry"s Law: k = Cg/Pg = (6.5 x 10-5 M)/(0.225 atm) = 0.00029 M/atm.

Under the new conditions the concentration would certainly be:

Cg = (0.00029 M/atm)(0.18 atm) = 5 x 10-5 M

The fish will have actually plenty of oxygen!

The vapor pressure over a pure solvent is 120 torr. ~ a non-volatile solute is added to the solvent the pressure over the systems is 80 torr. What is the mole portion of the solute in this solution?Answer

In this situation the solute is non-volatile and also so P°solute = 0.

Psoln = XsolventP°solvent + XsoluteP°solute = XsolventP°solvent

Xsolvent = Psoln / P°solvent = (80 torr)/(120 torr) = 0.667

Xsolvent + Xsolute = 1 ⇒ Xsolute = 1 - 0.667 = 0.333

Calculate the full vapor pressure above a equipment at 20 °C when 100 mole of C6H12 are combined with 10 mole C7H16. Right here is part data because that these building materials that deserve to be used in this problem and the next two problems:

P° (at 20°C)DensityMolec.Wt.
C6H1277.7 Torr0.779 g/mL84.16 g/mole
C7H1640 Torr0.684 g/mL100.21 g/mole
AnswerPtot = XC6H12P°C6H12 + XC7H16P°C7H16 = PC6H12 + PC7H16

 PC6H12 = ( 100 mole C6H12 110 Total Moles
) ( 77.7 Torr ) = 70.64 Torr (C6H12)

 PC7H16 = ( 10 mole C7H16 110 Total Moles
) ( 40 Torr ) = 3.636 Torr (C7H16)

PTotal = 74.28 TorrWhat is the vapor pressure over a solution made by combining 500 mL C6H12 and also 100 mL C7H16? watch the data table in the previous problem.Answer

Get the moles of each:

 ( 500 mL C6H12 ) ( 0.779 g 1 mL
)( 1 mole C6H12
84.16 g C6H12
) = 4.628 mole C6H12

 ( 100 mL C7H16 ) ( 0.684 g 1 mL
)( 1 mole C6H12
100.21 g C7H16
) = 0.6826 mole C7H16

 PC6H12 = ( 4.628 mole C6H12 5.311 Total Moles
) ( 77.7 Torr ) = 67.7 Torr (C6H12)

 PC7H16 = ( 0.6826 mole C7H16 5.311 Total Moles
) ( 40 Torr ) = 5.14 Torr (C7H16)

PTotal = 72.84 TorrWhat room the mole fractions of C6H12 and C7H16 if the full vapor pressure over the solution is 70 Torr? see the data table in the next to last problem.AnswerPtot = XC6H12P°C6H12 + XC7H16P°C7H16 ; 1 = XC6H12 + XC7H16

70 Torr = (1 - XC7H16)(77.7 Torr) + (XC7H16)(40 Torr)

Solve to get XC7H16 = 0.2 and also C6H12 = 0.8A mixture that sugar, C6H12O6, and 500 mL of water has a full vapor press of 9.0 torr in ~ 10 °C. How much street was added to the water? at 10 °C the vapor push of pure water is 9.21 torr and the density of water is 0.9997 g/mL.AnswerPtot = XC6H12O6P°C6H12O6 + XH2OP°H2O ; P°C6H12O6 = 0

XH2O = Ptot/P°H2O = (9 torr)/(9.21 torr) = 0.977 = (mole H2O)/(total moles)

 ( 500 mL H2O ) ( 0.997 g H2O 1 mL H2O
)( 1 mole H2O
18 g H2O
) = 27.77 mole H2O

⇒ 0.977 = (27.77 mole H2O)/(x mole C6H12O6 + 27.77 mole H2O)

Solve because that x to acquire 0.654 mole C6H12O6

(0.654 mole C6H12O6)(180 g/mole) = 118 g C6H12O6Describe the intermolecular pressures for each of the remedies (a), (b), and (c) that would reason the habits shown in each case.